In triangle ABC, m<A = 90, m<B=75, and BC = \sqrt{3}. What is the area of ABC? Express your answer as a common fraction.
HELP THIS IS SO HARD HHELP MEE!!! EXplain too please!
You have a right-angled triangle so we can use the formula
(1/2) base x height
Since you want the answer as a common fraction, I will assume that you need "exact" values.
you state that BC = \sqrt{3}
I will assume that is meant to say √3
so cos75 =base/√3
base = √3cos75
and sin75 = height/√3
height = √3sin75
area = (1/2)(√3sin75)(√3cos75)
= (3/2)sin75cos75)
(recall sin 2A = 2sinAcosA)
= (3/2)(2sin75cos75)/2
= (3/4) sin 150
sin 150 = sin30 ---- one of the basic identities
area = (1/3) sin150
= (3/4)sin30
= (3/4)(1/2) = 3/8
thanks alot
THANK YOU !!!!!!!
You don't even need trig for this. Reflect B, draw and altitude from B to B'C and solve
To find the area of triangle ABC, we can use the formula A = 1/2 * base * height. However, in this case, we don't have the height directly given.
Let's start by breaking down the information we have:
- We know that angle A is 90 degrees, which means triangle ABC is a right triangle.
- We also know that angle B is 75 degrees.
Since triangle ABC is a right triangle, we can use trigonometric ratios to find the length of the sides.
Using the sine function, we have sin(B) = Opposite / Hypotenuse, where B is angle B.
Here, the hypotenuse is BC, which is given as √3, and angle B is given as 75 degrees.
sin(75) = opposite / √3
To solve for the opposite side, let's multiply both sides by √3:
√3 * sin(75) = opposite
Now, we can find the value of √3 * sin(75) using a calculator:
√3 * sin(75) ≈ 2.8284
So, the length of the side opposite angle B is approximately 2.8284.
Now, we have the base (BC) and the height (opposite of angle B) of the triangle.
Area = 1/2 * base * height
= 1/2 * √3 * 2.8284
= (√3 * 2.8284)/2
= (2.8284 * √3)/2
= 1.4142 * √3
Thus, the area of triangle ABC is approximately 1.4142 * √3.